Question 1163738
<pre>
You did not draw the figure.  Ikleyn assumed the pipes were tied like this:

{{{drawing(170,170,-5,5,-5,5,

circle(2,2,2),circle(-2,2,2),circle(-2,-2,2),circle(2,-2,2),
line(-4,-2,-4,2),line(-2,-4,2,-4), line(4,-2,4,2),line(2,4,-2,4))}}}

Most people have had the experience of putting a rubber band tight around 4
identical pens, and have seen that they kind of shift on their own like the
figure below.  So I believe the figure below was what you were given, right?  

{{{drawing(200,200,-3,9,-3,9,
line(0,-2,4,-2),line(5.73205098,-1,7.73205098,-1+sqrt(12)),

line(-sqrt(3),1,2-sqrt(3),1+2sqrt(3)),line(2,2+sqrt(12),6,2+sqrt(12)),

circle(0,0,2.04), circle(4,0,2.04),circle(2,sqrt(12),2,04),circle(6,sqrt(12),2.04))}}}

I'll draw it bigger and put in some lines in red.

{{{drawing(600,600,-3,9,-3,9,
line(0,-2,4,-2),line(5.73205098,-1,7.73205098,-1+sqrt(12)),

line(-sqrt(3),1,2-sqrt(3),1+2sqrt(3)),line(2,2+sqrt(12),6,2+sqrt(12)),

circle(0,0,2), circle(4,0,2),circle(2,sqrt(12),2),circle(6,sqrt(12),2),


red(line(0,0,0,-2),line(0,0,4,0),line(4,0,4,-2),line(4,0,6,sqrt(12)), line(6,sqrt(12),6+sqrt(3),sqrt(12)-1), line(4,0,4+sqrt(3),-1),
line(6,sqrt(12),6,2+sqrt(12)),line(2,sqrt(12),2-sqrt(3),1+sqrt(12)),
line(2,sqrt(12),0,0),line(0,0,-sqrt(3),1),line(2,sqrt(12),2,2+sqrt(12)),
line(2,sqrt(12),6,sqrt(12)),line(2,sqrt(12),4,0)),

locate(1.3,3.6,90^o), locate(1.75,3.2,60^o),locate(2.3,3.5,60^o),
locate(2.11,4,90^o),

locate(5.3,3.47,60^o), locate(5.9,3.34,90^o),
locate(5.4,4,90^o)









)}}}

The four quadrilaterals which appear
to be rectangles indeed ARE rectangles because radii drawn
to the points of tangency are perpendicular to the tangent.
The interior angles of a rectangle are 90° each.

Therefore the 4 straight parts of the band are 2 radii each or
1 diameter or 12 inches,  So the straight parts of the band 
are 4∙12  = 48 inches. 

The triangle connecting the centers of any three mutually 
tangent circles is an equilateral triangle because the sides
are two radii or 1 diameter or 12 in. each.  The interior angles
of an equilateral triangle are 60° each.

So the 60° and 90° angles are correct as indicated in the 
figure.   

The circumference of any one of the 4 circles is &pi;∙d or 
12&pi; inches.

Adding the four marked angles in the upper left circle and 
subtracting from 360° gives 60° for each of the 2 shorter 
curved parts of the band.  Since 60° is 1/6 of 360°, each 
shorter curved part of the band is 1/6th of 12&pi; or 2&pi; 
inches  Therefore the 2 shorter curved parts of the band are 
4&pi; inches.

Adding the three marked angles in the upper right circle and 
subtracting from 360° gives 120° for each of the 2 longer 
curved parts of the band.  Since 120° is 1/3 of 360°, each 
longer curved part of the band is 1/3th of 12&pi; or 4&pi;  
Therefore the 2 longer curved parts of the band are 8&pi;
inches.

So the sum of all the curved parts is 4&pi;+8&pi; = 12&pi; inches.
Notice that they add up to one of the circle's circumference.

Adding the straight parts and the curved parts, we get

length of band = 48 + 12&pi; inches, about 85.68 inches.

So Ikleyn was right that the band has the same length whether
the pipes are arranged as Ikleyn assumed them to be or as I 
assumed them to be.    

Edwin</pre>